Naive Operations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others)Total Submission(s): 2114    Accepted Submission(s): 915 Problem Description In a galaxy far, far away, there are two integer sequence a and b of le…

Naive Operations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description In a galaxy far, far away, there are two integer sequence a and b of length…

Naive Operations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others)Total Submission(s): 1791    Accepted Submission(s): 772 Problem Description In a galaxy far, far away, there are two integer sequence a and b of le…

hdu 6315 题意:对于一个数列a,初始为0,每个a[ i ]对应一个b[i],只有在这个数字上加了b[i]次后,a[i]才会+1. 有q次操作,一种是个区间加1,一种是查询a的区间和. 思路:线段树,一开始没用lazy,TLE了,然后开始想lazy的标记,这棵线段树的特点是,每个节点维护 :一个区间某个a 要增加1所需个数的最小值,一个区间已加上的mx的最大值标记,还有就是区间和sum. (自己一开始没有想到mx标记,一度想把lazy传回去. (思路差一点就多开节点标记. #include…

Problem DescriptionIn a galaxy far, far away, there are two integer sequence a and b of length n.b is a static permutation of 1 to n. Initially a is filled with zeroes.There are two kind of operations:1. add l r: add one for al,al+1...ar2. query l r:…

题意:a数组初始全为0,b数组题目给你,有两种操作: 思路:dls的思路很妙啊,我们可以将a初始化为b,加一操作改为减一,然后我们维护一个最小值,一旦最小值为0,说明至少有一个ai > bi,那么找出所有为0的给他的最终结果加上一并且重置为bi,维护一个区间和,询问时线段树求和.一开始updateMin没加判断,单个复杂度飙到nlog(n),疯狂TLE... 代码: #include<cstdio> #include<vector> #include<stack>…

题意:数量为N的序列a和b,a初始全为0,b为给定的1-N的排列.有两种操作:1.将a序列区间[L,R]中的数全部+1:2.查询区间[L,R]中的 ∑⌊ai/bi⌋(向下取整) 分析:对于一个位置i,如果ai<bi,那么该位置不能对结果做出贡献:而当某一次操作后,ai>=bi了,就对结果的贡献值+1.那么可以用在线段树的结点中维护每个区间的最大a和最小b,和已经产生的贡献cnt.如果一个区间中最大的a超过了b,那么就说明此次更新操作使该区间的结果产生了变化,那么就要向下找到那个产生贡献的位置.…

发现每次区间加只能加1,最多全局加\(n\)次,这样的话,最后的答案是调和级数为\(nlogn\),我们每当答案加1的时候就单点加,最多加\(nlogn\)次,复杂度可以得当保证. 然后问题就是怎么判断答案是否该加1.我们可以用线段树设初值为给出的排列,把区间加改成区间减,维护最小值.当最小值为0是遍历左右子树,找到该加1的节点,一共会找\(nlongn\)次复杂度也可以得到保证. #include<iostream> #include<cstring> #include<c…

<题目链接> 题目大意: 给出两个序列,a序列全部初始化为0,b序列为输入值.然后有两种操作,add x y就是把a数组[x,y]区间内全部+1,query x y是查询[x,y]区间内∑[ai/bi].([ai/bi]代表ai/bi后向下取整) 解题分析: 首先,如果每次+1都暴力更新到每个叶子节点肯定会超时,但是如果不更新到叶子节点又不好维护每个节点对应区间 ∑[ai/bi] 的值,所以我们可以每个节点都维护四个值.sum值代表这个区间每个节点整数部分的所有数之和,lazy进行懒惰标记,避…

题目: http://acm.hdu.edu.cn/showproblem.php?pid=6315 Naive Operations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others) Problem Description In a galaxy far, far away, there are two integer sequence a and b of length…

6315.Naive Operations 题意很好理解,但是因为区间求和求的是向下取整的a[i]/b[i],所以直接分数更新区间是不对的,所以反过来直接当a[i]==b[i]的时候,线段树对应的位置更新+1操作是可取的,但是怎样才能在合适的时候+1操作呢?一开始智障想的是只要单点是b[i]的倍数就可以啊,但是这样就相当于单点查询的操作,铁定超时到上天,但是反过来就可以啊,直接一开始给一个数组赋值为b[i]的值,区间更新的时候所有的都更新,然后区间查询一下最小值,有0就说明有的已经正好减完b[i…

题目大意 题目链接Naive Operations 题目大意: 区间加1(在a数组中) 区间求ai/bi的和 ai初值全部为0,bi给出,且为n的排列,多组数据(<=5),n,q<=1e5 axmorz 思路 因为是整除,所以一次加法可以对ans 没有影响 当ai是bi的倍数,对ans会有贡献 所以我们维护一个sum,初值为bi(只对于线段树的叶子节点有用) 当区间+1的时候 我们对sum-1 当sum=0的时候(倍数) ans++,sum=bi 然后再维护一个区间最小值 当区间内的mi>…

Naive Operations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others)Total Submission(s): 2691    Accepted Submission(s): 1183 Problem Description In a galaxy far, far away, there are two integer sequence a and b of l…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6315 学习博客:https://blog.csdn.net/SunMoonVocano/article/details/81207676 Naive Operations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others)Total Submission(s): 4002    A…

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链接:HDU-6315:Naive Operations 题意: In a galaxy far, far away, there are two integer sequence a and b of length n.b is a static permutation of 1 to n. Initially a is filled with zeroes.There are two kind of operations:1. add l r: add one for al,al+1...a…

Naive Operations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others)Total Submission(s): 3636    Accepted Submission(s): 1612 Problem Description In a galaxy far, far away, there are two integer sequence a and b of l…

原题地址 Naive Operations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others)Total Submission(s): 887    Accepted Submission(s): 336 Problem Description In a galaxy far, far away, there are two integer sequence a and b o…

http://acm.hdu.edu.cn/showproblem.php?pid=6315 题意 a数组初始全为0,b数组为1-n的一个排列.q次操作,一种操作add给a[l...r]加1,另一种操作query查询Σfloor(ai/bi)(i=l...r). 分析 真的是太naive啦,现场时没做出来. 看见区间自然想起线段树,那么这里的关键就是整除问题,只有达到一定数量才会对区间和产生影响. 反过来想,先把a[i]置为b[i],那么每次add时就是-1操作,当a[i]为0时区间和+1,再把…

Problem Description In a galaxy far, far away, there are two integer sequence a and b of length n.b is a static permutation of 1 to n. Initially a is filled with zeroes.There are two kind of operations:1. add l r: add one for al,al+1...ar 2. query l…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6315 In a galaxy far, far away, there are two integer sequence a and b of length n. b is a static permutation of 1 to n. Initially a is filled with zeroes. There are two kind of operations: 1. add l r: a…

Four Operations Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 22    Accepted Submission(s): 12 Problem Description Little Ruins is a studious boy, recently he learned the four operations!Now h…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6315 题目大意:告诉你a,b两个数组,a数组初始化为0,b数组告诉你长度和具体值,接下来有q次操作,add操作是从向区间[l,r]加1,query操作是求区[l,r]的ai/bi的总和. 解题思路:维护一个mn,表示这个区间里的a最少加几次才能有新的贡献.比如1最多对总和造成q的贡献,2最多q/2,q/3,q/4,一直到q/n.最少要加min(b[i]-a[i]%b[i])次才能有新的贡献,如果这个…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4720 题目大意:给你四个点,用前三个点绘制一个最小的圆,而这三个点必须在圆上或者在圆内,判断最一个点如果在圆外则是安全的,否则是危险的. 解题思路:我是先借用了别人的模板求三个点组成的最小的圆的圆心以及半径,就列出了圆的标准方程,将第四个点代入其中,则即可以判断此点是否在圆上还是圆外. AC代码: #include <iostream> #include <cstdio> #inclu…

Naive and Silly Muggles Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 228 Accepted Submission(s): 163 Problem Description Three wizards are doing a experiment. To avoid from bothering, a special…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4720 解题报告:给出一个三角形的三个顶点坐标,要求用一个最小的圆将这个三个点都包含在内,另外输入一个点,问这个点是否在这个圆的圆外,如果在圆外,输出Safe,否则输出Danger. 现在的主要目的其实就是求这个最小的三角形的圆心坐标,很显然,当这个三角形是锐角三角形的时候,这个最小的圆就是这个锐角三角型的外接圆,否则就是以这个三角形最长的那条边的中点为圆心,以这条边的一半为半径的圆. #includ…

Naive and Silly Muggles Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 24    Accepted Submission(s): 17 Problem Description Three wizards are doing a experiment. To avoid from bothering, a spec…

题目链接 Problem Description In a galaxy far, far away, there are two integer sequence a and b of length n. b is a static permutation of 1 to n. Initially a is filled with zeroes. There are two kind of operations: 1.add l r: add one for $$$a_l, a_{l+1}..…

Problem Description Little Ruins is a studious boy, recently he learned the four operations! Now he want to use four operations to generate a number, he takes a string which only contains digits ‘1’ - ‘9’, and split it into 5 intervals and add the fo…

让ci = ai / bi, 求sum(ci)的值,因为每次 ai 都是加一的,那么我可以用一颗线段树来维护每个 i 位置的 ai 距离达到 bi 还需要的数的最小值,更新是每次都减一,如果我某一个区间的最小值等于 0, 这就说明我这时候的ai已经满足了ai/bi==1的情况,那么对应的ci的位置就应该加一,所以我每次发现一个区间的最小值是0的话,我就 dfs 去找到为0的地方,把这里重新修改成bi,然后这个位置的答案+1就可以了 #include<map> #include<set&g…